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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1001. |
The reduction potential values are given below `AI^(3_)//AI =- 1.67 "volt" Mg^(2+)//Mg =- 2.34 "volt" Cu^(2+)//Cu = +0.34` volt `I_(2)//2I^(-) = +0.53` volt. Which one is the best reducing agenet?A. `AI`B. `Mg`C. `Cu`D. `I_(2)` |
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Answer» Correct Answer - B Reducing agent, reduces other & get oxidized self higher negative value of `Mg^(2+)//Mg` indicates its tendency to get oxidized. |
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| 1002. |
From the values given in question No. 63, which one is the oxidising agent ?A. AlB. MgC. `I_(2)`D. Cu |
| Answer» Correct Answer - C | |
| 1003. |
Pick out the wrong statement, in electrochemical cellA. electrons are released at anodeB. cathode is regarded as negative electrodeC. chemical energy is converted into electrical energyD. salt bridge maintains the electrical neutrality of the solution. |
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Answer» Correct Answer - B In an electro-chemical cell cathode is positive electrode. |
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| 1004. |
The reduction potential values are given below `AI^(3+)//AI =- 1.67 "volt" Mg^(2+)//Mg =- 2.34 "volt" Cu^(2+)//Cu = +0.34` volt `I_(2)//2I^(-) = +0.53` volt. Which one is the best reducing agenet?A. `Al`B. `Mg`C. `Cu`D. `I_(2)` |
| Answer» Correct Answer - B | |
| 1005. |
Which one is the wrong statement about electrochemical series?A. Active metals have negative reduction potentialsB. Active non-metals have positive reduction potentialsC. Metals above hydrogen liberate hydrogen from acidsD. Metals below hydrogen are strong reducing agents |
| Answer» Correct Answer - D | |
| 1006. |
When a piece of sodium metal is dropped in water, a reaction takes place to yield hydrogen because:A. sodium loses electronsB. sodium acts as an oxidising agentC. water loses electronsD. water acts as a reducing agent |
| Answer» Correct Answer - A | |
| 1007. |
Which are true for a standard hydrogen electrode ?A. The hydrogen ion concentrated is 1MB. Temeprature is `25^(@)C`C. Pressure of hydrogen is 1 atmosphereD. It contains a metallic conductor which does not absorb hydrogen |
| Answer» Correct Answer - D | |
| 1008. |
For the half-cell reaction, `Au^(3+) + 3e^(-) rarr Au` the value of n used in Nernest equation is:A. 3B. 2C. 1D. `3 xx 96500` |
| Answer» Correct Answer - A | |
| 1009. |
Why does the conductivity of a solution decrease with dilution ? |
| Answer» The conductivity of a solution is related with the number of ions present per unit volume of the solution. When the solution is diluted, the number of ions decreases. Hence, conductivity or specific conductance of the solutiion also decreases. | |
| 1010. |
How would you determine the standard electrode potential of `Mg^(2+)|Mg`? |
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Answer» We will set up a cell consisting of `Mg|MgSO_(4)(1M)` as one electrode (by dipping a magnesium wire in 1M `MgSO_(4)` solution) and standard hydrogen electrode Pt, `H_(1)` (1 atm) `H^(+)|(1M)` as the second electrode and measure the EMF of the cell and also note the direction of deflection in the voltmeter. the direction of deflection shows that electronsf low from magnesium electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode. Hence, the cell may be represented as: `Mg|Mg^(2+)(1M)||H^(+)(1M)|H_(2),(1atm),Pt` `E_(cell)^(@)=E_(H^(+),1//2H_(2))^(@)-E_(Mg^(2+),Mg)^(@)` put `E_(H^(+),1//2H_(2))^(@)=0` Hence, `E_(Mg^(2+),Mg)^(@)=-E_(cell)^(@)`. |
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| 1011. |
How much electricity in terms of Coulomb is required to reduce 1 mol of Cr2O72- to Cr3+. |
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Answer» 2Cr2O7-2 → 2Cr+3, 2Cr+6 + 6e → 2Cr3+ Therefore the coulomb of electricity required = 6F, = 6 x 96500 C = 579000 C |
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| 1012. |
The concept of electrode potential is explained on the basis of :(a) Arrhenius’ theory (b) Ostwald’s theory (c) Nemst’s theory (d) Faraday’s law |
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Answer» Option : (c) Nemst’s theory |
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| 1013. |
The standard reduction potentials of metals A and B are x and y respectively. If x > y, the standard emf of the cell containing these electrodes would be :(a) 2x – y (b) y – x (c) x – y (d) x + y |
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Answer» Option : (c) x – y |
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| 1014. |
The emf of the cell,H2 | H+ || Cu2+ | Cu is :(1atm)(1M) (1M)(E0red = 0.34 V)(a) -1.34 (b) 0.34 V (c) -0.34 V (d) 1.34 |
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Answer» Option : (b) 0.34 V |
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| 1015. |
The Electromotive Force of the following Cell Cu|Cu++ (1 M)||A+g (1 M)|Ag is …………….. if \(E^0_{Cu^{++}}\) = 0.33 V and \(E^0_{Ag^{++}/Ag}\) = 0.79 V(a) 0.46 V (b) – 0.46 V (c) 1.12 V (d) – 112 V |
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Answer» Option : (a) 0.46 V |
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| 1016. |
The standard cell potential of the following cell is 0.463 V Cu|Cu++(1 M)||Ag+(1 M)|Ag. If E0Ag = 0.8 V, what is the standard potential of Cu electrode ?(a) 1.137 V (b) 0.337 V (c) 0.463 V (d) – 0.463 V |
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Answer» Option : (b) 0.337 V |
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| 1017. |
The metal which cannot displace hydrogen from dil. H2SO4 solution is :(a) Zn (b) Al (c) Fe (d) Ag |
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Answer» Option : (d) Ag |
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| 1018. |
With reference to the position of various metals in the electrochemical series, identify and justify the method of extraction followed for the metals given below : (i) Zinc (ii) Copper (iii) Magnesium |
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Answer» (i) Zinc is a metal present in the middle of the electrohemical series. Zinc has negative SRP value that is why it can be reduced by normal reducing agents such as hydrogen, carbon monoxide and carbon. This is because these reducing agents possess the ability to break ionic bonds between metal ion and oxide ion. (ii) Copper is a metal present above hydrogen in the electrochemical series. It has positive value for SRP that indicates the lesser thermal stability of its oxide. Thus, it can be extracted by reduction with other metals that have lower SRP than Cu. (iii) Magnesium is a metal present at the bottom in the electrochemical series. It has very high value of negative SRP value. That means, it itself is a strong reducing agent and hence, cannot be reduced by any other reducing agents. It is therefore extracted by the electrolytic reduction of its molten salt |
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| 1019. |
Which of the following is characteristic of the cathode in a voltaic cell?A. It may gain weight during reactionB. Electrons flow to it through the external circuitC. It is where oxidation occursD. it receives electrons from ions in solution |
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Answer» Correct Answer - A::B |
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| 1020. |
Define molar conductivity. What is the significance of it? |
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Answer» Molar conductivity : It is defined as a conductance of a volume of the solution containing ions from one mole of an electrolyte when placed between two parallel plate electrodes 1 cm apart and of large area, sufficient to accommodate the whole solution between them, at constant temperature. It is denoted by ∧m. Thus, The significance of molar conductivity is the conductance due to ions from one mole of an electrolyte. |
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| 1021. |
Obtain a relation between conductivity (κ) and molar conductivity (∧m). |
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Answer» Conductivity or specific conductance (κ) is the conductance of 1 cm3 of the solution in C.G.S. units, while molar conductivity is the conductance of a solution containing one mole of an electrolyte. Consider C molar solution, i.e., C moles of an electrolyte present in 1 litre or 1000 cm3 of the solution. ∴ C moles of an electrolyte are present in 1000 cm3 solution. ∴ 1 mole of an electrolyte is present in \(\frac{1000}{C}\) cm solution. Now, ∴ Conductance of 1 cm3 of this solution is κ, ∴ Conductance of \(\frac{1000}{C}\) cm3 of the solution is \(\frac{k\times 1000}{C}\) This represents molar conductivity, ∧m. ∴ ∧m = \(\frac{k\times 1000}{C}\)cm2mol-1(in C.G.S units) [ In case of SI units : Consider a solution in which C moles of an electrolyte are present in 1m3 of solution. Conductivity κ is the conductance of 1m3 of solution. ∵ C moles of an electrolyte are present in 1m3 solution. ∴ 1 mol of an electrolyte is present in \(\frac{1}{C}\) solution. ∵ Conductance of 1m3 of this solution is κ. ∴ Conductance of \(\frac{1}{C}\)m3 of the solution is \(\frac{k}{C}\). This represents molar conductivity, ∧m. ∴ ∧m = \(\frac{k}{C}\)Ω-1 m2 mol-1 (In SI units).] |
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| 1022. |
For the given cell, Mg|Mg2+|| Cu2+|Cu(i) Mg is cathode(ii) Cu is cathode(iii) The cell reaction is Mg + Cu2+ → Mg2+ + Cu(iv) Cu is the oxidising agent |
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Answer» (ii) Cu is cathode (iii) The cell reaction is Mg + Cu2+ → Mg2+ + Cu |
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| 1023. |
Molar conductivity of ionic solution depends on_____.A. temperatureB. distance between electrodesC. concentration of electrolysis in solutionD. surface area of electrodes. |
| Answer» Correct Answer - A::C::D | |
| 1024. |
In an electrolysis experiment, current was passed for `5h` through two cells connected in series. The first cell contains a solution of gold and second contains copper sulphate solution. In the first cell, `9.85g ` of gold was deposited. If the oxidation number of gold is `+3`, find the amount of copper deposited at the cathode of the second cell. Also calculate the magnitude of the current in ampere, `(` Atomic weight of `Au` is 197 and atomic weight of `Cu` is `63.5)`. |
| Answer» Correct Answer - 0.80A | |
| 1025. |
A `100-W,100-V` incardescent lamp is connected in series with an electrolyte cell containing cadmium sulphate solution. How much cadmium will be deposited by the current flowing for `10h` ? |
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Answer» Correct Answer - `19.06g` Watt`=`Ampere`xx`Volt `:.` Ampere `(I)=(Wat t)/(Vol t)=(100)/(110)` Now,`W=(ZIt)/(96500)` `:.W_(Cd)=(112.4xx100xx10xx60xx60)/(2xx110xx96500)` `=(404640000)/(21230000)=19.06g` |
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| 1026. |
A `100-W,110-V` incardescent lamp is connected in series with an electrolyte cell containing cadmium sulphate solution. How much cadmium will be deposited by the current flowing for `10h` ? |
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Answer» Current Strength `= ("Watt")/("Volt") = (100)/(110)` `= 0.909` ampere Given, `t = 10 xx 60 xx 60` sec, `E_(Cd) = (112.4)/(2)` `(Cd^(2+)+2e rarr Cd)` `w = (E.i.t)/(96500)` `w= (112.4 xx 0.909 xx 10 xx 60 xx 60)/(2 xx 96500) = 19.06 g` |
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| 1027. |
A current of `3.7A` is passed for `6h` between nickel electrodes in `0.5L` of a `2M` solution of `Ni(NO_(3))_(2)`. What will be the molarity of the solution at the end of electrolysis? |
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Answer» The electrolysis of `Ni(NO_(3))_(2)` in presence of `Ni` electrode will bring in following changes : At anode : `Ni rarr Ni^(2+)+2e` At cathode : `Ni^(2+)+2e rarr Ni` Eq. of `Ni^(2+)` formed = Eq. of `Ni^(2+)` lost Thus, there will be no change in conc. of `Ni(NO_(3))_(2)` sol. during electrolysis i.e., It will remain `2M`. |
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| 1028. |
Measurements show that the cell potential is affected by a number of factors such as (i) the nature of the substance that makes up each half-cell (ii) the concentration of dissolved ions and molecules (iii) the pressures of gases (iv) the terpreatureA. (i), (ii), (iii), (iv)B. (ii), (iv)C. (i), (ii)D. (ii), (iii), (iv) |
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Answer» Correct Answer - A These are the same factors that determine both the Gibbs energy difference between substances and the value of the equilibrium constant, `K`, for a chemical systeam. |
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| 1029. |
The `EMF` of a galvanic cell is measured byA. ammeterB. galvanometerC. voltmeterD. potentiometer |
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Answer» Correct Answer - D The capacity to do electical work by a cell is called lthe cell potential. It is expressed in volt `(V)`. We could use a voltmeter to measure the cell potenital but this would not give us the correct value of cell potential. The reason is that the cell potenital is dependent upon the concentreation of the elctrolyte which would change if we allow the current to flow in the circuit through the voltmeter. Hence, we must measuure the potential difference betweent he two half-cells when the cell is hald at almost constant composition and no current is flowing The potential differnce of a cell when no current is drawn out and when the cell is operating reversibly (i.e., very-2 slowly) is called the `emf` (electromotive force) of the cell. The measurement of the `emf` can by done by using a pote-tiometer. The potentioeter consits of a wire of uniform cross-section and high resistance so that negligible current is drawn out during the measurement. We make use of Weston cadmium cell (a commonly used standard cell) for the standardization) of a potentiometer. |
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| 1030. |
Knowing that `K_(sp)` for `AgCl` is `1.0 xx 10^(-10)`, calculate `E` for a silver `//` silver chloride electrode immersed in `1.00 M KCl ` at `25^(@)C.E^(c-)._(Ag^(o+)|Ag)=0.799V`. |
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Answer» `K_(sp)(AgCl)=[Ag^(o+)][Cl^(c-)],KCl=1.00M` `:. [Cl^(c-)]=1.00M` `:. [Ag^(o+)]=(K_(sp))/([Cl^(c-)])=(1.0xx10^(-10))/(1.00)=1.0=10^(-10)` `Ag^(o+)+e^(-)rarr Ag" "E^(c-)._(Ag^(o+)|Ag)=0.799V` `:. E=E^(c-)-(0.059)/(1)log.(1)/([Ag^(o+)])` `=0.799-(0.059)/(1)log.(1)/((0.10xx10^(-10))` `=0.799=0.592=0.207V` |
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| 1031. |
The `EMF` of a galvanic cell is measured byA. voltmeterB. potentiometerC. galvanometerD. ammeter . |
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Answer» Correct Answer - B EMF is measured by potentiometer . |
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| 1032. |
The cell reaction `Zn(s)+2Ag^(+)(aq)rarrZn^(2+)(aq)+Ag(s)` is best represented byA. `Ag|Ag^(+)||Zn|Zn^(2+)`B. `Zn|Zn^(2+)||Ag^(+)|Ag`C. `2Ag|Ag^(+)||Zn|Zn^(2+)`D. `Zn|Zn^(2+)||2Ag|Ag^(+)` |
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Answer» Correct Answer - B For the cell reaction `Zn(s)+2Ag^(+)(aq)rarr2Ag(s)+Zn^(2+)(aq)` The two half cell reactions are `Zn(s)rarrZn^(2+)(aq)+2e^(-)` `Ag^(+)(aq)+e^(-)rarrAg(s)xx2` `Zn(s)+2Ag(aq)rarrZn^(2+)+2Ag(s)` Therefore Zn is oxidised and as such acts as -ve electrode or anode. `Ag^(+)` ions get reduced as such Ag electrode acts as cathode or _ve electrode. Therefore cell representation is `Zn|Zn^(2+)||Ag^(+)|Ag` |
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| 1033. |
The cell reaction `Zn(s)+2Ag^(+)(aq)rarrZn^(2+)(aq)+Ag(s)` is best represented byA. `Ag |Ag^(+)|| Zn|Zn^(2+)`B. `Zn | Zn^(2+) ||Ag^(+) | Ag`C. `2Ag |Ag^(+) || Zn|Zn^(2+)`D. `Zn|Zn^(2+) ||2Ag |Ag^(+)` |
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Answer» Correct Answer - B Electrode on which oxidation occurs is written on L.H.S and the other on the R.H.S. as represented by (b) . |
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| 1034. |
Aluminium displaces hydrogen from acids, but copper does not. A galvanic cell prepared by combining `Cu//Cu^(2+)` and `Al//Al^(3+)` has an emf of 2.0V at 298. if the potential of copper electrode is +0.34V. That of Aluminium electrode isA. `-2.3V`B. `+2.34V`C. `-1.66V`D. `1.66V`. |
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Answer» Correct Answer - C Cell representation is `Al|Al^(2+)||C u^(2+)|Cu` `E_("cell")=E_(Cu^(2+))+_(//Cu)-E_(Al^(3))+_(//Al)` `2.0V== 0.34V-E_(Al^(3+)//Al)` `E_(Al^(3+)//Al)=0.34V-2.0V=-1.66V` |
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| 1035. |
A fuel cell isA. the voltaic cells in which discontinuous supply of fuels are send at anode to give oxidationB. the voltaic cell in which fuels such as : `CH_(4) , H_(2) , CO` are used up at anodeC. it involves the reactions of `H_(2) - O_(2)` fuel cell such as Cathode : `2 H_(2) + 4OH^(-) to 4 H_(2) O(l) + 4e` Anode : `O_(2) + 2 H_(2)O (l) + 4e to 4 OH^(-)`D. the efficiency of a `H_(2) - O_(2)` fuel cell is very low |
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Answer» Correct Answer - B `CH_(4) , H_(2) , CO` are used fuel . |
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| 1036. |
Electrical potential of a cell is anA. intensive propertyB. extensive propertyC. isothermal propertyD. isobaric property. |
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Answer» Correct Answer - A Electrical potential of a cell is an intensive property. |
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| 1037. |
Which incorrect about fuel cells ?A. Cells continuously run as long as fuels are suppliedB. These are more efficient and free from pollutionC. These are used to provide power and drinking water to astronauts in space programmeD. Fuel cell has low efficiency |
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Answer» Correct Answer - D Efficiency very high . |
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| 1038. |
Which of the following is a catalyst used in fuel cellA. HgB. AgC. FeD. Pb |
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Answer» Correct Answer - B In fuel cell , Ag . Pt , Pd are used as a catalyst . |
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| 1039. |
Calculate `E^(C-)` for each cell and write the equation for each cell precess. Explaing the significance of any negative. `E^(c-)` value. `a.` `Cd(s)|Cd^(2+)(1.0M)||AgNO_(3)(1.0m)|Ag(s)` b. `Fe(s)|FeSO_(4)(1.0M)||ZnSO_(4)(1.0m)|Zn(s)` `c.` `Pt, Cl_(2)(g)(1 atm)|NaCl(1.0M)|Hg_(2)Cl_(2)(s)|Hg(s)` Given `E^(c-)._(Cd)=-0.40V,E^(c-)._((Fe))=-0.41V,E^(c-)._((Zn))=-0.76V` `E^(c-)._((Ag))=+0.80V,E^(c-)._((2Cl^(c-)|Cl))=-1.36V.` `E^(c-)._((Hg|Hg_(2)Cl_(2)))`=-0.27V.` |
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Answer» `a.` Anode `:" "Cd(s) rarr Cd^(2+)(1.0M)+ `cancel(2e^(-))` Cathode `:` `2Ag^(o+)(1.0M)+ `cancel(2e^(-))` rarr 2Ag(s) Cell reaction `:` `ul(bar(Cd(s)+2Ag^(o+)(1.0M)rarrCd^(2+)(1.0M)+2Ag(s)))` `E^(c-)._(cell)=(E^(c-)._(reduction ))_(cathode)-(E^(c-)._(reduction))_(anode )` `=0.80-(-0.4)=1.2V` `EMF` of cell is positive, hence the cell will function with `Cd` as anoe and `A` as cathode. `b.` Anode `: Fe(d) rarr Fe^(2+)(1.0M)+ cancel(2e^(-))` Cathode`: Zn^(2+)(1.0M)+ cancel(2e^(-)) rarr Zn(s)` Cell reaction `:` `ulbar(Fe(s)+Zn^(2+)(1.0M)rarrFe^(2+)(1.0M)+Zn(s))` `E^(c-)._(cell)=(E^(c-)._(reduction))_(cathode)-(E^(c-)._(reduction))_(anode)` `=0.76-(-0.41)=-0.35V`. Negative `EMF` valuse suggests that cell will not function in the manner it is represented , `i.e.,Fe` as anode and `Zn` as cathode. So reversing `(` interchanging `)` the cathode and anode, `i.e,` making `Zn` as anode and `Fe` as cathode, can make the cell functional. `c.` Anode `: cancel(2Cl)(1.0M)rarrCl_(2)(1.0atm)+cancel(2e^(-))` Cathode`:` `Hg_(2)Cl_(2)(s)+cancel(2e^(-)) rarr 2Hg(s)+2Cl^(c-)(1.0M)` Cell reaction `:` `ulbar(Hg_(2)Cl_(2)(s)rarr 2Hg(s)=Cl_(2)(1.0atm))` `E^(c-)._(cell)=(E^(c-)._(reduction))_(cathode )-(E^(c-)._(reduction))_(anode)` `=0.27-(1.36)=-1.09V` Negative `EMG` value suggests that the cell will note function in the manner it is represented, `i.e., HgCl_(2)` as cathode and `2Cl^(c-)|Cl_(2)` as anode. So reversing`(` interchanging `)` the cathode and anode, `i.e.,` making `Hg_(2)Cl_(2)( i.e., Hg|Hg_(2)Cl_(2))` as anode and `Cl_(2)|2Cl^(c-)` as cathode, can make the cell functional. |
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| 1040. |
Anode reaction of a fuel cell is:-A. `Zn(Hg)+2OH^(-)toZnO_((s))+H_(2)O+2e^(-)`B. `Pb_((s))+SO_(4(aq))^(2-)toPbSO_(4(s))+2e^(-)`C. `2H_(2(g))+4OH_((aq))^(-)to4H_(2)O_((l))+4e^(-)`D. `2Fe_((s))to2Fe^(2+)+4e^(-)` |
| Answer» Correct Answer - C | |
| 1041. |
Which of the following is NOT used as a fuel in fuel cellA. NitrogenB. HydrogenC. Carbon DioxideD. Methane |
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Answer» Correct Answer - A In fuel cell except `N_(2)` all the other gases are used. . |
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| 1042. |
How would you determine the standard reduction potential of the system `Mg^(2+)|Mg`? |
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Answer» A cell will be set up consisting of `Mg//MgSO_(4)(1M)` as one electrode and standard hydrogen electrode Pt, H, (1atm) `H^(+)//(1M)` as second electrode, measure the EMF of the cell and also note the direction of deflection in the voltmeter. The direction of deflection shows that `e^(-1)` s flow from mg electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reductio on hydrogen electrode. Hence, the cell may be represetned as follows. `Mg|Mg^(2+)(1M)||H^(+)(1M)|H_(2),(1atm),Pt` `E_(Cell)^(@)=E_(H^(+)//(1)/(2)H_(2))^(@)-E_(Mg^(2+)//Mg)^(@)` Put `E_(H^(+)//(1)/(2)H_(2))^(@)=0` `thereforeE_(Mg^(2+)//Mg)^(@)=-E_(cell)^(@)` |
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| 1043. |
Which of the following statements is correct for a galvanic cell?A. Reduction occurs at cathodeB. Oxidation occurs at anodeC. Electrons flow from anode to cathodeD. All the statements are correct . |
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Answer» Correct Answer - D All the first three statements are correct . |
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| 1044. |
If `E^(c-)._((Ag|Ag^(o+)))=-0.8V` and `E^(c-)._((H_(2)|2H^(o+)))=0 V` , in a cell arrangement using these two electrodes, find `E^(c-)._(cell)` and find out which electrode acts as anode and which acts as cathode. |
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Answer» Note that `E^(c-)._((Ag|Ag^(o+)))` and `E^(c-)._(H_(2)|2H^(o+))` are stanard oxidation potential values. Comparing the two values, `E^(c-)._((H_(2)|2H^(o+)))gtE^(c-)._((Ag|Ag^(o+)))` or Comparing the reduction potential values, `E^(c-)._((Ag^(o+)|Ag))gtE^(c-)._((2H^(o+)|H_(2)))` Hence, silver electrode will act as catohde and hydrogen electrode will act as anode. The cell representation of the cell is `:` `Pt,H_(2)(1 atm)|H^(o+)(1.0M)||Ag^(o+)(1.0M)|Ag(s)` `E^(c-)._(cell)=(E^(c)._(reduction))_(cathode)-(E^(c-)._(reduction))_(anode)` `=0.8-0=0.8V` |
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| 1045. |
The reaction at anode in a fuel cell isA. `Pb + SO_(4)^(2-) to Pb SO_(4) + 2 e^(-)`B. `2H_(2) + 4 OH^(-) to 4 H_(2)O +4e^(-)`C. `O_(2) + 2 H_(2)O + 4e^(-) to 4OH^(-)`D. `2H_(2) + O_(2) to 2H_(2)O` |
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Answer» Correct Answer - B It is a fact . |
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| 1046. |
Point out the correct statement about `Zn-CuSO_(4)` cell.A. The flow of electrons occurs from copper to zinc .B. The value of `E_("Red")^(@)` of copper electrode is less then that of zinc electrode .C. Zinc is anode while Cu is cathode electrode .D. All the statements are correct . |
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Answer» Correct Answer - C In `Zn - CuSO_(4)` cell , zinc is anode while copper is cathode . |
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| 1047. |
Which of the following is not a function of salt bridge ?A. To allow the flow of cations from one solution to the otherB. To allow the flow of anions from one solution to the otherC. To allow the electrons to flow from one solution to the otherD. To maintain electrical neutrality of the two solutions . |
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Answer» Correct Answer - C Salt bridge allows ions to flow through it . |
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| 1048. |
The EMF of a Daniel cellA. 0 voltB. 5 voltsC. 1.1 voltsD. 3 volts |
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Answer» Correct Answer - C Emf of Daniel cell is 1.1 V . |
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| 1049. |
EMF of a cell is equal to the potential difference between the two electrodes when the current flowing in the circuit is_____. |
| Answer» Correct Answer - Zero | |
| 1050. |
In preparing a salt bridge we sue `KCl` becauseA. `K^(+)` and `Cl^(-)` are iso-electronicB. `K^(+)` and `Cl^(-)` have the same transport numberC. KCl is a strong electrolyteD. KCl forms a good jelly with agar-agar. |
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Answer» Correct Answer - B KCl is preferred in a salt bridge because `K^(+)` and `Cl^(-)` ions have same transport number . |
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